# scalene triangle

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16 years ago
#1
how do i find the area of a scalene triangle! im sooo stuck please help. the perimeter has got to equal 1000m
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16 years ago
#2
Originally posted by *tashb*
how do i find the area of a scalene triangle! im sooo stuck please help. the perimeter has got to equal 1000m
Need more info - what are the angles? what are the length of the sides? otherwise the area could be pretty much anything
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16 years ago
#3
You can use 1/2 ab sin C

Where a and b are the sides that make angle C.

That is assuming you are given these pieces of data.

Otherwise you can use trig to calculate the vertical height and use 1/2 base x height.

Depends what data you are given.
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16 years ago
#4
(Original post by *tashb*)
how do i find the area of a scalene triangle! im sooo stuck please help. the perimeter has got to equal 1000m
Use Herons Formel. s=(a+b+c)/2

area=sqr(s(s-a)(s-b)(s-c))

Regards EvilED
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16 years ago
#5
I've been trying to find the formula for solving the angles of a scalene triangle.... any help???
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16 years ago
#6
(Original post by Unregistered)
I've been trying to find the formula for solving the angles of a scalene triangle.... any help???
½a·b·sinC
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16 years ago
#7
A simply way to do this is to draw the triangle in a box so that every point in the triangle touches the box. Then you will have a few right angled triangles to work out, and take that away from the area of the box....

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16 years ago
#8
(Original post by Unregistered)
I've been trying to find the formula for solving the angles of a scalene triangle.... any help???
Cosine Rule and Sin Rule were devised for scalene triangles.
Cosine rule a, b, c are sides and A, B, C are the vertecices opposite their lower cased letter.
a^2 = b^2 + c^2 -2bccos A
b^2 = a^2 + c^2 -2accos B
c^2 = a^2 + b^2 -2abcos B

Using same notation

a/sinA=b/sinB=c/sinC {unknown lengths}
or the inverse
sinA/a=sinB/b=sinC/c {unknown angles}

Use which ever version puts your unknown as the numerator.
Maybe you will notice that right angled triangles when used with these formulae show formulae we are familiar with:
Cos 90 = 0
so in Cosine rule in A was 90deg
a^2 = b^2 + c^2 -2bccos A
becomes
a^2 = b^2 + c^2 ie Pythagoras theorum

Sin 90 = 1
so in Sine rule in A was 90deg
a/sinA=b/sinB
Becomes
a=b/sinB
rearranges to
sinB = b/a ie Opp over Hyp basic right angles Trig
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